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3y+(2-4y)=2-(5y-12)
We move all terms to the left:
3y+(2-4y)-(2-(5y-12))=0
We add all the numbers together, and all the variables
3y+(-4y+2)-(2-(5y-12))=0
We get rid of parentheses
3y-4y-(2-(5y-12))+2=0
We calculate terms in parentheses: -(2-(5y-12)), so:We add all the numbers together, and all the variables
2-(5y-12)
determiningTheFunctionDomain -(5y-12)+2
We get rid of parentheses
-5y+12+2
We add all the numbers together, and all the variables
-5y+14
Back to the equation:
-(-5y+14)
-1y-(-5y+14)+2=0
We get rid of parentheses
-1y+5y-14+2=0
We add all the numbers together, and all the variables
4y-12=0
We move all terms containing y to the left, all other terms to the right
4y=12
y=12/4
y=3
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